So let's assume that $V$ has a non-degenerate bilinear form $\langle\cdot,\cdot\rangle$ with a basis $e_1,\dots,e_n$ such that $\langle e_i,e_j\rangle = \delta_{ij}$, the Kronecker delta. Let $*$ denote the Hodge star operator. Note that we have the formula
$$ \langle x,y\rangle = *((*x)\wedge y) .$$
Let's identify any operator on $V$ with its matrix ...

As by the question linked by Michael, it is generally proved in a course about differential forms that $\alpha\wedge\beta = (-1)^{pq} \beta\wedge\alpha$ where $p$ and $q$ are the degrees of $\alpha$, $\beta$. Using this the result is immediate.
But you could try and prove it, at least in your specific case. Hint: it should not be difficult to prove it for ...